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m11#9 : Retired Discussions [Locked]

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I think the explanation is missing/incomplete. Please help me through this problem.

Which of the following sets must have the same standard deviation as set {a, b, c}?

A. {ab, b^2, cb} B. {2a, b + a, c + b} C. {0, b + a, c - a} D. {ab, bc, ac} E. {ab + c, a(1 + b), b(1+a)}

If we add or subtract a constant to each term in a set the standard deviation will not change.

Notice that set {(ab + c, a(1 + b), b(1+a)}={c+ab, a+ab, b+ab}, so this set is obtained by adding some number ab to each term of set {a, b, c}, which means that those sets must have the same standard deviation.

The standard deviation of a set does not change if a constant is added to all the members.

Thus, standard deviation of (a,b,c) will be the same as of (a+ab, b+ab, c+ab).

And, option E is the same as (a+ab, b+ab, c+ab).

Beautiful approach by scthakur. Thats the best approach to this question. +1.

SD of a, b and c and (a+x), (b+x) and (c + x) is the same. Trying to find exactly what is the SD of a, b and c and the same of each of the options in the question doesnot help solve this question. What helps is understanding the question. _________________

My approach was actually using real numbers such as a=2, b=3 and c=4. Though abit lengthy it worked since I applied the rule, the less spread out my answers were the closer my answer was making it E. Thanx now I know another rule;The standard deviation of a set does not change if a constant is added to all the members.

I took a similar, although longer, approach to solving this problem as the person above me. Immediately understanding that this problem was evaluating the spread, I relied on the use of "plugging" numbers in for a,b,c (1,2,3) and then looked for a similar spread amongst the answer choices.

Having read and followed the Manhattan Advanced Quant books, I first started with E and realized that this is the right answer --> matches to my "target"

Would of been even quicker if I would of realized that "ab" is consistent, a constant, throughout the 3 terms; and adding a constant to the terms does not alter the spread. Thanks for the clarification on this one guys!